顺时针打印矩阵

leetcode54

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]

You should return [1,2,3,6,9,8,7,4,5].

 

public List<Integer> spiralOrder(int[][] matrix) {
  List<Integer> res = new ArrayList<>();
  if (matrix.length == 0)
    return res;
  int rowBegin = 0;
  int rowEnd = matrix.length - 1;
  int colBegin = 0;
  int colEnd = matrix[0].length - 1;
  while (rowBegin <= rowEnd && colBegin <= colEnd) { //外圈遍历完，遍历内圈
    for (int j = colBegin; j <= colEnd; j++) //向右遍历
      res.add(matrix[rowBegin][j]);
    rowBegin++;

    for (int j = rowBegin; j <= rowEnd; j++)//向下遍历
      res.add(matrix[j][colEnd]);
    colEnd--;

    if (rowBegin <= rowEnd) {
      for (int j = colEnd; j >= colBegin; j--)//向左遍历
        res.add(matrix[rowEnd][j]);
    }
    rowEnd--;

    if (colBegin <= colEnd) {
      for (int j = rowEnd; j >= rowBegin; j--)//向上遍历
        res.add(matrix[j][colBegin]);
    }
    colBegin++;
  }

  return res;
}